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I'm trying to figure this out, and I feel I'm pretty close to why it's the case. I just can't quite get the details to work.

Let $X$ be an adapted process on $(\Omega,\mathcal{F},\mathbb{P})$ and $T$ a finite stopping time. Show that $X_T$ is $\mathcal{F}$-measurable.

As I understand it, the process $X_T = \{X^T_n\}_{n\geq1}$ is defined as $X^T_n(\omega) := X_{T(\omega) \wedge n}(\omega)$. Since $X$ is an adapted process we have that $X_n$ is $\mathcal{F}_n$-measurable for all $n$, and since $\mathcal{F}_n \subset \mathcal{F}$ this intuitively should carry over to $X_T$. The $\mathcal{F}_n's$ are from the filtration of $X$. My problem is with $T(\omega)$ being dependant on $\omega$, and that $T$ isn't necessarily bounded which means you can't bound $T(\omega)\wedge n$.

Any tips on how I should approach this?

EDIT: some clarification

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You should explain the notations: what is $\mathcal F$? I also think the process is discrete (by your attempt), but you have to tell it. Furthermore, I think a $n$ is missing in the definition of $X_T^n(\omega)$. –  Davide Giraudo Oct 21 '12 at 20:00
    
Your definitions are ambiguous: usually, $X_T$ is not a process but the random variable $X_T:\omega\mapsto X_{T(\omega)}(\omega)$. The process $(X_{T\wedge n})_n$ is often abbreviated as $X^T$. –  Did Oct 21 '12 at 21:05
    
Ah, I see what you mean. I'm still getting to terms with all the notation, it can be a little confusing. –  BallzofFury Oct 21 '12 at 21:08
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2 Answers 2

up vote 3 down vote accepted

Hint Show that everything in the RHS of the formula below is measurable with respect to $\mathcal F$: $$ X_T=\sum_n\mathbf 1_{T=n}\cdot X_n $$ Note If really the object of interest is the process $X^T=(X_{T\wedge n})_n$, use the formula $$ X^T_n=X_{T\wedge n}=\mathbf 1_{T\geqslant n}\cdot X_n+\sum_{k\lt n}\mathbf 1_{T=k}\cdot X_k $$ and show that everything in the RHS is measurable with respect to $\mathcal F_n$.

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I finally had some time to look at your reply properly. It seems I was looking at the wrong definition of $X_T$. Since $\{T=n\} \in \mathcal{F}_n$ by the definition of $T$, we have that $1_{\{T=n\}}$ is measurable. And since $X_n$ is also per definition $\mathcal{F}_n$-measurable, the product is also $\mathcal{F}_n$-measurable. Since $\mathcal{F}_n \subset \mathcal{F}$ for all $n \in \mathbb{N}$, we have that $X_T$ is $\mathcal{F}$-measurable. –  BallzofFury Oct 22 '12 at 18:01
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The idea is to decompose over the sets on which we know $T(\omega)$. Write $$\small (X^T_n)^{-1}(B)=\bigcup_{k\geq 0}(X_{k\wedge n}^{-1}(B)\cap \{\omega\mid T(\omega)=k\})=\bigcup_{k=0}^{n-1}(X_k^{-1}(B)\cap\{\omega\mid T(\omega)=k\}) \cup (X_n^{-1}(B)\cap \{\omega\mid T(\omega)\leq n\}^c)\in\mathcal F_n.$$

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